An electron microscope is operated at 40 kV. | vedclass

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(B) The de Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda_e = \frac{12.27}{\sqrt{V}} \ \mathring{A} = \fra

Vedclass · Accepted Answer

$9.78 	imes 10^4$

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(B) The de Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda_e = \frac{12.27}{\sqrt{V}} \ \mathring{A} = \frac{12.27}{\sqrt{40 	imes 10^3}} 	imes 10^{-10} \ m$.Calculating this,we get $\lambda_e \approx 6.135 	imes 10^{-12} \ m$.The resolving power $R$ of a microscope is inversely proportional to the wavelength of the radiation used,i.e.,$R \propto \frac{1}{\lambda}$.Therefore,the ratio of the resolving power of the electron microscope $(R_1)$ to the optical microscope $(R_2)$ is $\frac{R_1}{R_2} = \frac{\lambda_2}{\lambda_1}$.Substituting the values: $\frac{R_1}{R_2} = \frac{6 	imes 10^{-7}}{6.135 	imes 10^{-12}} \approx 0.978 	imes 10^5 = 9.78 	imes 10^4$.

An electron microscope is operated at $40 \ kV$. The ratio of the resolving power of this microscope to another one which uses yellow light of wavelength $6 \times 10^{-7} \ m$ is:

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